A=( $\frac{2}{x+1}$ +$\frac{1}{1-x}$ -$\frac{5}{1-x^2}$ ):$\frac{1-2x}{x^2-1}$ Rút gọn A

Đáp án:

`A=(\frac{2}{x+1}+\frac{1}{1-x}-\frac{5}{1-x^{2}}):\frac{1-2x}{x^{2}-1}` `(đk: x\ne\pm1;x\ne1/2)`

`=[\frac{2.(x-1)}{(x-1).(x+1)}-\frac{1.(x+1)}{(x-1).(x+1)}+\frac{5}{(x-1).(x+1)}].\frac{x^{2}-1}{1-2x}`

`=[\frac{2x-2-x-1+5}{(x-1).(x+1)}].\frac{x^{2}-1}{1-2x}`

`=\frac{x+2}{x^{2}-1}.\frac{x^{2}-1}{1-2x}`

`=\frac{(x+2).(x^{2}-1)}{(x^{2}-1).(1-2x)}`

`=\frac{x+2}{1-2x}`

Vậy `A=\frac{x+2}{1-2x}` với `x\ne\pm1;x\ne1/2`